Question: Simplify and expand the following expression: $ \dfrac{1}{p + 6}- \dfrac{4}{4p + 4}- \dfrac{4p}{p^2 + 7p + 6} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{4}{4p + 4} = \dfrac{4}{4(p + 1)}$ We can factor the quadratic in the third term: $ \dfrac{4p}{p^2 + 7p + 6} = \dfrac{4p}{(p + 6)(p + 1)}$ Now we have: $ \dfrac{1}{p + 6}- \dfrac{4}{4(p + 1)}- \dfrac{4p}{(p + 6)(p + 1)} $ The least common multiple of the denominators is: $ (p + 6)(p + 1)$ In order to get the first term over $(p + 6)(p + 1)$ , multiply by $\dfrac{4(p + 1)}{4(p + 1)}$ $ \dfrac{1}{p + 6} \times \dfrac{4(p + 1)}{4(p + 1)} = \dfrac{4(p + 1)}{(p + 6)(p + 1)} $ In order to get the second term over $(p + 6)(p + 1)$ , multiply by $\dfrac{p + 6}{p + 6}$ $ \dfrac{4}{4(p + 1)} \times \dfrac{p + 6}{p + 6} = \dfrac{4(p + 6)}{(p + 6)(p + 1)} $ In order to get the third term over $(p + 6)(p + 1)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4p}{(p + 6)(p + 1)} \times \dfrac{4}{4} = \dfrac{16p}{(p + 6)(p + 1)} $ Now we have: $ \dfrac{4(p + 1)}{(p + 6)(p + 1)} - \dfrac{4(p + 6)}{(p + 6)(p + 1)} - \dfrac{16p}{(p + 6)(p + 1)} $ $ = \dfrac{ 4(p + 1) - 4(p + 6) - 16p} {(p + 6)(p + 1)} $ Expand: $ = \dfrac{4p + 4 - 4p - 24 - 16p}{4p^2 + 28p + 24} $ $ = \dfrac{-20 - 16p}{4p^2 + 28p + 24}$ Simplify: $ = \dfrac{-5 - 4p}{p^2 + 7p + 6}$